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\title{\heiti\zihao{2} 习题17.2}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{计算 $I=\int\limits_{L}\left(x^{2}+y^{2}\right) \mathrm{d} x$,其中 $L$ 为椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1(y \geqslant 0)$ 上从点 $A(a, 0)$ 到点$B(-a, 0)$ 的一段弧.}
\textbf{解}\quad
令$x=a\cos\theta,y=b\sin\theta,x'(t)=-a\sin\theta$.
$$
\begin{aligned}
    I=\int\limits_{L}\left(x^{2}+y^{2}\right) \mathrm{d} x&=\int_0^\pi (a^2\cos^2\theta+b^2\sin^2\theta)\cdot -a\sin\theta\mathrm{d}\theta\\
    &=a\int_0^\pi a^2\cos^2\theta+b^2(1-\cos^2\theta)\mathrm{d}\cos\theta\\
    &=a\int_1^{-1}a^2u^2+b^2-b^2u^2\mathrm{d}u\\
    &=a\left(\dfrac{2(b^2-a^2)}{3}\right)-2ab^2\\
    &=\dfrac{2a(b^2-a^2)}{3}-2ab^2
\end{aligned}
$$


\section{计算 $I=\int\limits_{L}(x+y) \mathrm{d} x$, 其中 $L$ 是以点 $O(-1,0), A(1,0)$ 和 $B(0,1)$ 为顶点的三角形的边界,逆时针方向.}
\textbf{解}\quad
由于积分微元是$x$,所以其为第二型曲线积分.由第二型曲线积分的性质,可将积分曲线拆为三段:$L_1:(-1,0)--(1,0)$, $L_2:(0,1)--(1,0)$, $L_3:(-1,0)--(0,1)$.在这三条曲线上,分别有
$$
\begin{array}{ccc}
    y=0 & y = 1 - x & y = 1 + x
\end{array}
$$
所以有
$$
\begin{aligned}
    I&=\int\limits_{L1}+\int\limits_{L2}+\int\limits_{L3}(x+y)\mathrm{d}x\\
    &=\int_{-1}^1x\mathrm{d}x+\int_1^01\mathrm{d}x+\int_{0}^{-1}1+2x\mathrm{d}x\\
    &=-1
\end{aligned}
$$

\section{计算 $I=\int\limits_{L} x \mathrm{~d} y$, 其中 $L$ 为圆周 $x^{2}+y^{2}=a^{2}$ 与两个坐标轴围成的在第一象限区域的边界,沿顺时针方向.}
\textbf{解}\quad
令$x=a\cos\theta,y=a\sin\theta.$
$$
\begin{aligned}
    \int\limits_{L} x \mathrm{~d} y&=\int_{\pi/2}^{0}a\cos\theta a\cos\theta\mathrm{d}\theta\\
    &=-\dfrac{a^2\pi}{4}
\end{aligned}
$$

\section{计算 $I=\int\limits_{L} y \mathrm{~d} x+z \mathrm{~d} y+x \mathrm{~d} z$, 其中 $L$ 由从 $(2,0,0)$ 到 $(3,4,5)$ 的线段 $L_{1}$ 和从 $(3,4,5)$ 到 $(3,4,0)$ 的竖直线段 $L_{2}$ 组成.}
\textbf{解}\quad
$L_1:x=2+t, y = 4t, z = 5t,t\in(0,1)$.
$$
\begin{aligned}
    \int\limits_{L1}y\mathrm{d}x+z\mathrm{d}y+x\mathrm{d}z&=\int_0^1(4t+20t+10+5t)\mathrm{d}t =\dfrac{49}{2}\\
    \int\limits_{L2}y\mathrm{d}x+z\mathrm{d}y+x\mathrm{d}z&= \int_5^03\mathrm{d}z = -15
\end{aligned}
$$
从而
$$
\int\limits_{L2}y\mathrm{d}x+z\mathrm{d}y+x\mathrm{d}z = \dfrac{19}{2}
$$

\section{计算 $\int\limits_{L} y \mathrm{~d} x+z \mathrm{~d} y+x \mathrm{~d} z$, 其中 $L$ 是平面 $x+y=2$ 和球面 $x^{2}+y^{2}+z^{2}=2(x+y)$ 交成的圆周,从原点向曲线看去,该曲线取顺时针方向.}
\textbf{解}\quad
$(x-1)^2+(y-1)^2+z^2=2, x+y=2$.将$y=2-x$代入可得$z^2=4x-2x^2$.则$L$可由两个隐函数表示,为
$$
\left\{\begin{array}{l}
    x\\
    y = x\\
    z = \pm\sqrt{4x-2x^2}
\end{array}\right.
$$
从而
$$
\begin{aligned}
    \int\limits_{L}y \mathrm{~d} x+z \mathrm{~d} y+x \mathrm{~d} z&=\int\limits_{L1}y \mathrm{~d} x+z \mathrm{~d} y+x \mathrm{~d} z+\int\limits_{L2}y \mathrm{~d} x+z \mathrm{~d} y+x \mathrm{~d} z\\
    &=\int_2^0(2-x)\mathrm{d}x-\sqrt{4x-2x^2}\mathrm{d}y+x\dfrac{2-2x}{\sqrt{4x-2x^2}}\mathrm{d}x\\
    &+\int_0^2(2-x)\mathrm{d}x+\sqrt{4x-2x^2}\mathrm{d}x-x\dfrac{2-2x}{\sqrt{4x-2x^2}}\mathrm{d}x\\
    &=2\sqrt{2}\pi
\end{aligned}
$$
\section{计算 $\int\limits_{L}\left(y^{2}-z^{2}\right) \mathrm{d} x+\left(z^{2}-x^{2}\right) \mathrm{d} y+\left(x^{2}-y^{2}\right) \mathrm{d} z$, 其中 $L$ 为球面片 $x^{2}+y^{2}+z^{2}=1$,$x \geqslant 0, y \geqslant 0, z \geqslant 0$ 的边界,从第一卦限向原点看去,该曲线取逆时针方向.}
\textbf{解}\quad
可将曲线划分为在三个平面上的三个部分,记$L1,L2,L3$分别为在$xoy,yoz,zox$上的曲线.由于三个部分的积分关于$x,y,z$轮换对称,从而有
$$
\begin{aligned}
    \int\limits_{L}\left(y^{2}-z^{2}\right) \mathrm{d} x+\left(z^{2}-x^{2}\right) \mathrm{d} y+\left(x^{2}-y^{2}\right) \mathrm{d} z&=3\int\limits_{L1}\left(y^{2}-z^{2}\right) \mathrm{d} x+\left(z^{2}-x^{2}\right) \mathrm{d} y+\left(x^{2}-y^{2}\right) \mathrm{d} z\\
    &=3\int\limits_{L1}y^2\mathrm{d}x-x^2\mathrm{d}y\\
    &=3\int_0^{\pi/2}-\sin^3\theta - \cos^3\theta\mathrm{d}\theta\\
    &=-4
\end{aligned}
$$

\section{计算第二型曲线积分$$I=\oint\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+y^{2}}$$}
\subsection{$L: x^{2}+y^{2}=a^{2}$, 取逆时针方向}
\textbf{解}\quad
极坐标变换
$$
\begin{aligned}
    \oint\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+y^{2}}&=\int_0^{2\pi}\dfrac{a^2\cos^2\theta+a^2\sin^2\theta}{a^2}\mathrm{d}\theta\\
    &=\int_0^{2\pi}\mathrm{d}\theta = 2\pi
\end{aligned}
$$


\subsection{$L:|x| \leqslant 1,|y| \leqslant 1$ 的边界,取逆时针方向.}
\textbf{解}\quad
令$L1,L2,L3,L4$为$$(1,-1)->(1,1),(1,1)->(-1,1),(-1,1)->(-1,-1),(-1,-1)->(1,-1)$$
$$
\begin{aligned}
    \oint\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+y^{2}}&=\oint\limits_{L1}\dfrac{\mathrm{d}y}{1+y^2}-\oint\limits_{L2}\dfrac{\mathrm{d}x}{1+x^2}-\oint\limits_{L3}\dfrac{\mathrm{d}y}{1+y^2}+\oint\limits_{L4}\dfrac{\mathrm{d}x}{1+x^2}\\
    &=\int_{-1}^1\dfrac{\mathrm{d}y}{1+y^2}-\int_1^{-1}\dfrac{\mathrm{d}x}{1+x^2}-\int_1^{-1}\dfrac{\mathrm{d}y}{1+y^2}+\int_{-1}^{1}\dfrac{\mathrm{d}x}{1+x^2}\\
    &=2\pi
\end{aligned}
$$


\section{设 $P, Q$ 为长度为 $l$ 的光滑曲线 $L$ 上的连续函数,证明:}
\subsection{$\left|\int\limits_{L} P(x, y) \mathrm{d} x+Q(x, y) \mathrm{d} y\right| \leqslant M \cdot l$, 其中 $M=\max\limits_{(x, y) \in \mathrm{L}} \sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}} ;$}
\begin{proof}
    $|\int\limits_L\boldsymbol{f}\mathrm{d}\boldsymbol{s}|\leqslant|\int\limits_L||\boldsymbol{f}||\mathrm{d}s|\leqslant M\int\limits_L\mathrm{d}s=M\cdot l$
\end{proof}


\subsection{利用这个不等式估计 $I_{R}=\oint\limits_{x^{2}+y^{2}=R^{2}} \dfrac{y \mathrm{~d} x-x \mathrm{~d} y}{\left(x^{2}+x y+y^{2}\right)^{2}}$, 并证明 $\lim\limits_{R \rightarrow \infty} I_{R}=0 .$}

\begin{proof}$$
\begin{aligned}
    P=\dfrac{y}{(x^2+xy+y^2)^2}\\
    Q=\dfrac{-x}{(x^2+xy+y^2)^2}\\
    \sqrt{P^2+Q^2}=\dfrac{\sqrt{x^2+y^2}}{(x^2+xy+y^2)^2}\leqslant\dfrac{2R}{R^4}=\dfrac{2}{R^3}
\end{aligned}
$$
所以
    $$
        \begin{aligned}
            0\leqslant\oint\limits_{x^{2}+y^{2}=R^{2}} \dfrac{y \mathrm{~d} x-x \mathrm{~d} y}{\left(x^{2}+x y+y^{2}\right)^{2}}\leqslant\dfrac{4\pi R}{R^3}=\dfrac{4\pi}{R^2}
        \end{aligned}
    $$
    从而
    $$
    \lim\limits_{R \rightarrow \infty} I_{R}=\lim\limits_{R \rightarrow \infty}\dfrac{4\pi}{R^2}
    $$
\end{proof}

\section{设质点在变力 $\boldsymbol{F}(x, y)=\left(1+y^{2}, 2 x+y\right)$ 的作用下,从点 $O(0,0)$ 沿曲线 $y=a \sin x$
$(a>0)$ 运动到 $A(\pi, 0)$}
\subsection{求变力 $\boldsymbol{F}(x, y)$ 所做的功}
\textbf{解}\quad
$$
\begin{aligned}
    P=1+y^2\\
    Q=2x+y
\end{aligned}
$$
$$
\begin{aligned}
    \int\limits_LP\mathrm{d}x+Q\mathrm{d}y&=\int_0^\pi(1+a^2\sin^2x+2x+a\sin x)a\cos x\mathrm{d}x\\
    &=\dfrac{\pi(2+a^2)}{2}-4a
\end{aligned}
$$
\subsection{当 $a$ 为何值时变力 $\boldsymbol{F}(x, y)$ 所做的功最小}
\textbf{解}\quad
二次函数$\dfrac{\pi}{2}a^2-4a+\pi$在$a=\dfrac{4}{\pi}$时取最小值$\pi=\dfrac{8}{\pi}$


\end{document}